Isomorphism between two exact sequences

Question

I am trying to understand a proof and came across the following diagram (for any abelian category): $$$$

$\hskip2in$

The proof now says:
Since the rows of the diagram are exact, there are unique dashed arrows making the diagram commute.

Question:
Is there a common lemma or reason that I am missing on why there are unique $A \rightarrow A'$ and $ D \rightarrow D'$ making the diagram commute?

The goal would be to have isomorphisms $A \rightarrow A'$ and $ D \rightarrow D'$ which would be the case by simply swapping the rows and doing the same again as $B \rightarrow B'$ and $C \rightarrow C'$ are isomorphisms.

Answer 1

The composition $A\to B\to C$ is zero, therefore so is $A\to B\to C\to C'$. But that's the same as $A\to B\to B'\to C'$. But $A'\to B'$ is the kernel of $B'\to C'$. Therefore $A\to B\to B'$ factors uniquely through $A'\to B'$. That is, there is a unique arrow making the left square commute.

For the right square, dualise.